Is it true that the only interesting topologies are metric topologies and weak topologies? - MathOverflow

Is it true that the only interesting topologies are metric topologies and weak topologies?

2010-04-16T13:16:10Z

In "Infinite dimensional analysis, A hitchhikers guide" by Aliprantis and Border, they write that these 2 classes of topologies "by and large include everything of interest".

@Pete Clarke: I was asking if it holds for mathematics in general, and not just for functional analysis. From the answers I get the impression that their statement is a fairly good first approximation.

@Gerald Edgar: I thought I read in Mathoverflow that the Zariski topology can be regarded as a weak topology (at least in some cases).

Answer by Joel David Hamkins for Is it true that the only interesting topologies are metric topologies and weak topologies?

2010-04-16T13:24:50Z

Every topological space X has the initial topology (or weak topology) with respect to the family of continuous functions from X to the Sierpiński space. (see http://en.wikipedia.org/wiki/Initial_topology.)

This is the two point space {a,b} with open sets: emptyset, {a} and {a,b} only. If U is any subset of X, then the function f_U, mapping points in U to a and the rest to b, will be continuous if and only if U is open.

Answer by Andrew Stacey for Is it true that the only interesting topologies are metric topologies and weak topologies?

2010-04-16T13:38:23Z

Picking up on Gerald's interpretation of the question (namely, that it really focusses on infinite dimensional vector spaces) then I say: absolutely not!

For example, piecewise-smooth paths in some Euclidean space has a topology that is neither of these (it's an uncountable inductive limit of Frechet spaces). (Not that I particularly recommend this space!)

Close to Gerald's comment, "dual-Frechet" spaces (that is, the dual of a Frechet space with the strong topology) have very nice properties, almost as nice as Frechet spaces themselves. This class includes distributions with the "right" topology.

And that's the point, really. If you're only interested in, say, distributions for what they can say about compactly supported functions, then the weak topology is probably fine. However, if you are interested in distributions in their own right then the weak topology is very unlikely to be okay.

Here's an example from my research: I like infinite dimensional manifolds, and I quite like loop spaces. To construct the Dirac operator on a loop space, I needed to put an inner product on the cotangent bundle. So I needed, in effect, a continuous injective map $(L\mathbb{R}^n)^* \to H$ ($H$ being some standard Hilbert space). I can't do this with the weak topology any continuous map from $(L\mathbb{R}^n)^*$ with the weak topology to a normed vector space has to factor through a finite dimensional space. With the strong topology, though, it was no problem.

So, deal with metric and weak topologies if you like; but real analysts use the strong topology[1].

[1] Not sure what complex analysts use.

Answer by Johannes Hahn for Is it true that the only interesting topologies are metric topologies and weak topologies?

2010-04-16T14:02:20Z

Frechet spaces, limits of Frechet spaces were mentioned before. I'd like to emphasize a particularly important example (which was also mentioned before, but I want to extend it a bit): The space of test functions $\mathcal{D}(\Omega)$ is a strong inductive limit of Frechet spaces, neither metrizable nor do they carry the weak topology. These spaces are the foundation of distribution theory and therefore most important.

on the other hand the other two usual spaces of test function $\mathcal{E}(\Omega)$ and the schwartz space $\mathcal{S}(\mathbb{R}^n)$ are metrizable because they can be topologized by a countable family of (semi)norms.

The distribution space $\mathcal{D}'(\Omega), \mathcal{E}'(\Omega), \mathcal{S}'(\mathbb{R}^n)$ can be endowed with the weak topology (that is the pointwise convergence). But the strong topology is also common and this is again neither weak nor metrizable.

Answer by Matthew Daws for Is it true that the only interesting topologies are metric topologies and weak topologies?

2010-04-18T10:11:20Z

Google book's has access to the book in question:

http://books.google.com/books?id=4hIq6ExH7NoC&lpg=PP1&ots=p8sOXwh3Ny&dq=Aliprantis%20and%20Border&pg=PA47#v=onepage&q=by%20and%20large%20include%20everything%20of%20interest&f=false

So the meaning of "weak topology" is that which Joel defines. This was actually a new idea to me, but it seems very general (more general that what I would have called the "weak topology" in Functional Analysis). As Joel points out, interpreted in maximal generality, any topological space carries the weak topology in this sense.

The book actually very quickly specialises to weak topologies generated by continuous, real-valued functions. Then your space has to be completely regular. I think it would be fair to say that this does rule out some interesting examples.

Answer by Andrej Bauer for Is it true that the only interesting topologies are metric topologies and weak topologies?

2010-04-18T11:58:25Z

Of course, any question of the form "Is it the case that the things considered interesting by such-and-such small set of people are all the interesting things?" is going to have a negative answer. Worse, taking statements such as yours too literally and outside of appropriate context will just cause you to be closed-minded and unable to understand other people's ideas.

Even if you are interested only in functional analysis you will quickly hit against topologies which are not included among your topologies. For example, given an infinite-dimensional Hilbert space $H$, what is a good topology to put on its lattice of closed subspaces? (This topology won't even be Hausdorff.)

For a second example, consider how we reconstruct a compact Hausdorff space $X$ from its $C^{*}$-algebra $A = C(X,\mathbb{C})$ of continuous complex maps. As the points of the reconstructed space we take the maximal ideals in $A$, and declare that the closed sets are the ideals of $A$ (please correct me if I am getting closed/open and ideal/filter wrong here, I always do). Even though at the end of the day this topology turns out to be metrizable, that is entirely orthogonal to how the topology is defined and thought of.